Let $f:\left[4,\infty \right)\to \left[1,\infty \right)$ be a function defined by $f\left(x\right)=5^{x\left(x-4\right)}$ , then $f\left(x\right)=5^{x\left(x-4\right)}$ is

Let $f^{-1}\left(x\right)=y$  $\Rightarrow x=f\left(y\right)$

               $x=5^{y\left(y-4\right)}$

               ${\log }_5^x=y^2-4y$

               $y^2-4y-{\log }_5^x=0$

             $y=f^{-1}\left(x\right)=2+\sqrt{4+{\log }_5^x}$